On the Hughes’ model for pedestrian flow: The

one-dimensional case

Marco Di Francescoa, Peter A. Markowichb,c, Jan-Frederik Pietschmannb,Marie-Therese Wolframb

aDipartimento di Matematica Pura ed Applicata, Universita degli Studi dell’Aquila, Via VetoioLoc. Coppito, 67100 L’Aquila, Italy

bDepartment of Applied Mathematics and Theoretical Physics (DAMTP), University ofCambridge. Wilberforce Road, Cambridge CB3 0WA, UK

cFaculty of Mathematics, University of Vienna, Nordbergstrasse 15, 1090 Vienna, Austria

Abstract

In this paper we investigate the mathematical theory of Hughes’ model for the flowof pedestrians (cf. [17]), consisting of a nonlinear conservation law for the densityof pedestrians coupled with an eikonal equation for a potential modelling the com-mon sense of the task. For such an approximated system we prove existence anduniqueness of entropy solutions (in one space dimension) in the sense of Kruzkov[22], in which the boundary conditions are posed following the approach of Bardoset al. [7]. We use BV estimates on the density ρ and stability estimates on thepotential φ in order to prove uniqueness. Furthermore, we analyse the evolutionof characteristics for the original Hughes’ model in one space dimension and studythe behaviour of simple solutions, in order to reproduce interesting phenomena re-lated to the formation of shocks and rarefaction waves. The characteristic calculusis supported by numerical simulations.

Keywords: Pedestrian flow; Scalar conservation laws; Eikonal equation; Ellipticcoupling; Entropy solutions; Characteristics.

1. Introduction

The mathematical modelling of large human crowds has gained a lot of scien-tific interest in the last decades. This is due to various reasons. First of all, a veryserious issue in this context is to shed a light on the dynamics in critical circum-stances. A well known practical example is the Jamarat Bridge in Saudi Arabia:the huge number of pilgrims cramming the bridge on occasion of the pilgrimage toMecca gave rise to serious pedestrian disasters in the nineties [15]. Moreover, theanalytical and numerical study of the qualitative behaviour of human individuals

Preprint submitted to Journal of Differential Equations September 20, 2010

in a crowd with high densities can improve traditional socio–biological investiga-tion methods. The dynamics of a human crowd has also applications in structuralengineering and architecture: the London Millennium Footbridge which had to beclosed on the day of its opening due to unexpected anomalous synchronization,is a very evocative example in this sense. Other applications of pedestrian flowmodelling arise in transport systems, spectator occasions, political demonstrations,panic situations such as earthquakes and fire escapes. More light-hearted exam-ples are the simulation of pedestrian movement in computer games and animatedmovies, see [38].

Several models for the movement of crowds have been proposed in the past.One can distinguish between two general approaches: microscopic and macroscopicmodels. In the microscopic framework, people are treated as individual entities(particles). The evolution of the particles in time is determined by physical andsocial laws which describe the interaction among the particles as well as theirinteractions with the physical surrounding. Examples for microscopic methodsare social-force models (see [14] and the references therein), cellular automata, e.g[12, 30], queuing models e.g. [40] or continuum dynamic approaches like [38]. Foran extensive review on different microscopic approaches we refer to [13]. Notethat the microscopic approach in [38] uses the eikonal equation to compute thepedestrians’ optimal path. This is a common feature with the model we willanalyse in this paper.

In contrast to microscopic models, macroscopic models treat the whole crowdas an entity without considering the movement of single individuals. Classicalapproaches use well known concepts from fluid and gas dynamics, see [16]. Morerecent models are based on optimal transportation methods [29], mean field games(see [24] for a general introduction) or non-linear conservation laws [8]. In [32],an approach based on time-evolving measures is presented. We finally note thatcrowd motion models share many features with traffic models [1].

In this paper we shall analyse a model introduced by R. L. Hughes in 2002[17]. Hughes’ model treats the crowds as a “thinking” fluid and has been appliedto diverse scenarios like the Battle of Agincourt and the annual Muslim Hajji [18].It is given by

ρt − div(ρf 2(ρ)∇φ) = 0 (1a)

|∇φ| = 1

f(ρ)(1b)

Here x denotes the position variable with x ∈ Ω, a bounded domain in Rd withsmooth boundary ∂Ω, t ≥ 0 is time and ρ = ρ(x, t) is the crowd density. Thefunction f(ρ) is given by f(ρ) = 1 − ρ, modelling the existence of a maximaldensity of individuals which can be normalized to 1 by a simple scaling. System

2

(1) is supplemented with the following boundary conditions for φ

φ(x, t) = 0, x ∈ ∂Ω, t ≥ 0 (2)

and the initial conditionρ(x, 0) = ρI(x) ≥ 0. (3)

We shall be more precise about the boundary conditions for ρ and give a moredetailed interpretation of the model in the next section.

Note that if the term 1f(ρ)

in (1b) is replaced by 1, the system decouples and

(1a) reduces to a non-linear conservation law with discontinuous flux. This typeof equation has been analysed and simulated in [19, 20]. Even though Hughes’system (1) shares some features with this class of equations it is methodologicallymuch more challenging. This is due to the non-linearity of the eikonal equation(1b) as well as the implicit time dependence of the potential ∇φ in (1a). In fact,for the unique viscosity solution φ of the eikonal equation, no more regularitythan Lipschitz continuity can be expected. In this paper we present an existenceand uniqueness theory for a regularized version of (1) in one space dimension.Additionally, we discuss the behaviour of simple solution for the original system(1) and validate these results numerically.

Numerical simulations are already available in literature, see Ling et al. [27].Their approach does not cover the case of discontinuous flux inside the computa-tional domain. Nevertheless we follow the iterative procedure presented in [27],i.e. first solve the eikonal equation (1b) then the conservation law (1a). Nu-merical methods for non-linear conservation laws with discontinuous flux can befound in literature, e.g. [37]. We will use the approach presented by J. Towersfor our numerical simulations. Note that equation (1a) is similar to the Lighthill-Witham-Richards traffic flow model [26, 34], and similar numerical schemes can beused. Various approaches can be found in the literature, e.g. [6, 5, 41, 42]. Theseschemes are usually based on numerical methods for non-linear conservation laws,for a general introduction we refer to [25, 36] and the references therein.

This paper is organized as follows: In the remaining part of the introduction,we shall explain the model in more detail (subsec. 1.1), present regularized versions(subsec. 1.2) and state our main results (subsec. 1.3). In sec. 2, we prove existenceand uniqueness of entropy solutions for a regularized model and in sec. 3 we willanalyse some special cases for the non regularized problem and compare the resultswith our numerical simulations.

3

1.1. Hughes’ model

We start with a brief motivation of Hughes’ model (1) (for further details see[17]). The density of individuals ρ = ρ(x, t) satisfies the continuity equation

ρt + div(ρV ) = 0, (4)

and we use the following ‘polar decomposition’ notation for the velocity field V (x, t)

V (x, t) = |V (x, t)|Z(x, t), |Z(x, t)| = 1. (5)

In order to prescribe a logistic dependency of |V | with respect to ρ we choose theclassical linear expression

|V (x, t)| = 1− ρ.

As for the directional unit vector Z(x, t), we assume it to be parallel to the gra-dient of the potential φ(x, t). Such potential is determined by solving the eikonalequation in (1). The potential φ rules the common sense of the task (the task isrepresented by the boundary ∂Ω). More precisely, the pedestrians tend to mini-mize their estimated travel time to the target. In a very naive way, this could bemodelled by prescribing the eikonal equation

|∇φ| = 1, φ|∂Ω = 0,

which has the unique semi-concave solution φ(x) = dist(x, ∂Ω) at least in the caseof a convex domain Ω. However, it is reasonable to assume that individuals tempertheir estimated travel time by avoiding extremely high densities, i. e.

|∇φ| = 1

1− ρ, φ|∂Ω = 0, (6)

which implies a ‘density driven’ rearrangement of the level sets of φ. This leadsto Z(x, t) = ∇φ(x,t)

|∇φ(x,t)| = (1 − ρ)∇φ and therefore the continuity equation in (1) isjustified.

1.2. An attempt to a mathematical theory: approximations

A successful attempt to develop a mathematical theory for the model (1) hasnever been carried out so far. The non-linearity with respect to ρ in the continuityequation forces using the notion of entropy solution for scalar conservation laws, asit is well known that weak L∞ solutions to such kind of equations are in general notunique. On the other hand, the vector field ∇φ may clearly develop discontinuitiesin subsets of Ω which may vary in time.

In general, the subsets of discontinuity of∇φ depend on ρ non-linearly and non–locally. This may be seen by simple examples in one space dimension. Moreover,

4

the presence of the term 1 − ρ in the right-hand-side of the eikonal equationrenders the problems even more difficult, because of the possible blow–up of |∇φ|as ρ approaches the over-crowding density ρ = 1.

A full understanding of the model is highly non-trivial, even in one spacedimension, where the model can be decoupled by solving the eikonal equation byintegration.

In order to overcome such difficulties, we propose reasonable approximationsto the Hughes’ model (1), basically consisting of a regularization of the potentialto avoid the discontinuity of |∇φ|. At a first glance, a very natural way to approx-imate the equation for the potential would be simply adding a small ‘viscosity’, i.e.

−δ∆φ+ |∇φ|2 =1

f(ρ)2, δ > 0.

Such an approximation still has the drawback of (possibly) producing a blow upof the right hand side when the density approaches the overcrowding value ρ = 1.This problem can be bypassed considering instead

−δ∆φ+ f(ρ)2|∇φ|2 = 1, δ > 0. (7)

On the other hand, the development of a satisfactory existence and uniquenesstheory by using the coupling (7) is seriously complicated by the presence of thedensity dependent coefficient multiplying the Hamilton-Jacobi term |∇φ|2.

The model for which we shall develop a full existence and uniqueness theoryuses the following elliptic regularization of the eikonal equation in (1), namely

−δ1∆φ+ |∇φ|2 =1

(f(ρ) + δ2)2, δ1, δ2 > 0. (8)

The sign in front of δ1 (δ in the alternative equation (7)) is chosen such thatwe would recover the unique viscosity solution in a possible limit δ1 → 0. Thesecond order term in (8) is meant to smooth the potential φ in order to avoiddiscontinuities for |∇φ|. The elliptic operator in (8) is a classical elliptic Hamilton-Jacobi operator, and it is therefore easier to deal with if compared to the one in(7). On the other hand equation (8) contains one further approximation on theright-hand-side which can be motivated as follows.

Without the elliptic regularization, the potential φ in (8) would satisfy

|∇φ| = 1

(1− ρ+ δ2)(9)

5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Figure 1: Comparison between the scalar ‘logistic’ speed |V | of the pedestrian in Hughes’ model(1) (left) and the model with elliptic coupling (8)

Then, the polar decomposition of the velocity field introduced in (4) reads in thiscase

V = |V |Z, |Z| = 1

|V | = f(ρ)2|∇φ| = f(ρ)2

δ2 + f(ρ)=

(1− ρ)2

δ2 + (1− ρ), Z =

∇φ|∇φ|

. (10)

The profile of |V | as a function of ρ in (10) has essentially the same properties ofthe logistic function |V |(ρ) = 1 − ρ of the original Hughes’s model, except thatthe vacuum at ρ = 1 is achieved with a zero derivative and the maximal velocityis slightly penalized, i. e. |V |max = 1/1 + δ2 instead of |V |max = 1 of the originalmodel (cf. Figure 1).

As for the unit vector Z, which is parallel to ∇φ, the only difference withthe original model is that individuals ‘sense’ the target as the density reaches themaximum value ρ = 1. In this case |∇φ| = 1/δ2, i. e. the slope of ∇φ is very highin absolute value (δ2 is thought as a small parameter), but not infinite as in theoriginal model. On the other hand, when ρ = 1, |V | vanishes, and therefore theabove mentioned difference is not effective (individuals do not move at all whenρ = 1!).

1.3. Results

We shall first cover the one dimensional existence and uniqueness theory for theregularized model with elliptic coupling (8) introduced in the previous subsection,more precisely we shall study the model systemρt − (ρf 2(ρ)φx)x = 0

−δ1φxx + |φx|2 =1

(f(ρ) + δ2)2.

(11)

6

As the continuity equation in (11) features non-linear convection, we shalladdress the existence and uniqueness theory in the framework of weak entropysolutions, cf. for instance [22]. The results are contained in Section 2. Moreprecisely, the notion of solution is stated in Definition 2.1, the existence result isprovided in Theorem 2.10, and the uniqueness result is proven in Theorem 2.12.

The problem (11) is posed on the bounded interval x ∈ [−1, 1] with homoge-neous Dirichlet boundary conditions. We shall follow the approach by Bardos etal. [7] (see also [9, 2, 28]) to recover suitable boundary conditions for a scalarconservation law. This aspect is explained at the beginning of the next section.

2. The regularized model: existence and uniqueness theory

In this section we establish our existence and uniqueness results for the regu-larized Hughes’ model system (11) with f(ρ) = (1− ρ). For future use we denote

g(ρ) := ρf(ρ)2.

System (11) is coupled with the initial condition

ρ(x, 0) = ρI(x) ≥ 0, (12)

and with the Dirichlet boundary conditions

mink∈[0,tr ρ]

g(tr (ρ))− g(k) = 0, (13)

φ(±1, t) = 0. (14)

Here tr ρ denotes the trace of ρ on the boundary. More precisely,

tr ρ(−1, t) = limx→−1+

ρ(x, t), tr ρ(1, t) = limx→1−

ρ(x, t).

It was originally proven in [7] that (13) is the correct way to pose Dirichlet bound-ary conditions for a scalar conservation law, mainly for two reasons: first, (13)comes as a natural condition from the vanishing viscosity limit of solutions withzero Dirichlet boundary data; second, (13) encloses the natural interplay betweenthe boundary datum and the value of the solution which is transported via char-acteristics in the linear case (the boundary datum needs to be posed only if char-acteristics at the boundary are directed towards the interior of the domain). Theboundary condition provided in [7] assumes the simplified form (13) since we shalldeal with non-negative solutions and due to a trivial monotonicity property of thepotential φ (cf. Lemma 2.11 below). We remark here that the boundary condition(13) reduces to

g(tr ρ) ≥ g(k) on x = ±1, for all k ∈ [0, tr ρ],

7

which expresses the fact that the allowed densities on the boundary are those forwhich the function g is non-decreasing. A deeper understanding of the boundaryconditions for nonlinear conservation laws in one space dimension can be also foundin [9].

We shall prove that the system (11) has a unique solution (ρ, φ) in a sensemade precise by the following definition. For the density component ρ we will usethe classical notion of entropy solutions originally introduced by Kruzkov in [22]and adapted to boundary value problems by Bardos et al. in [7].

Definition 2.1 (Entropy Solution). Let ρI ∈ BV ([−1, 1]). A couple (ρ, φ) is aweak entropy solution to the system (11) if

• ρ ∈ BV ([−1, 1]× [0, T )) ∩ L∞([−1, 1]× [0, T ))

• φ ∈ W 2,∞[−1, 1]

• ρ and φ satisfy the inequality∫∫ΩT

|ρ− k|ψt dxdt+

∫ ∞−∞

ρIψ0dx−∫∫ΩT

sgn(ρ− k)[g(ρ)− g(k)]ψxφx dxdt

+

∫∫ΩT

sgn(ρ− k)g(k)ψφxxdxdt− sgn(k)

∫ T

0

[g(tr ρ)− g(k)]φxψ|x=±1dt ≥ 0,

(15)

for every Lipschitz continuous test function ψ on [−1, 1]× [0, T ) having com-pact support.

• φ and ρ satisfy the second equation in (11) almost everywhere in x and t.

As usual in the context of conservation laws, we shall approximate the targetedmodel (11) via a vanishing viscosity approach, namely we shall work on the system

ρt − (ρf 2(ρ)φx)x = ερxx (16a)

−δ1φxx + |φx|2 =1

(f(ρ) + δ2)2, (16b)

for a small ε > 0. System (16) is coupled with homogeneous boundary condition

ρ(x, t)|x=±1 = 0 φ(x, t)|x=±1 = 0,

and the initial conditionρ(x, 0) = ρI(x).

8

Existence of unique (smooth) solutions to the above regularized problem followfrom standard results. For the elliptic coupling see e.g. [23, Chapter 3, Lemma 1.1]and [23, Chapter 3, Thm. 1.2]. For the parabolic approximation we refer to [39,Section 5, Thm. 5.3 and Thm. 5.4]. The proof of this theorem is based on semigroup techniques. The strategy is to first linearise the equation to an evolutionequation with a linear but time depending operator. Under the given assumptions,it is known that there exists a solution to such an equation (see e.g. [35]). Then,the solution to the non-linear equation in obtained using a fixed-point argument.

In the next subsections we shall first derive suitable a-priori estimates on φ andρ, then we shall recall our notion of entropy solution, and finally prove existenceand uniqueness of the limit as ε→ 0.

2.1. A Priori Estimates on φ and ρ

We shall now derive some a-priori estimates for the elliptic coupling, i.e.

− δ1φxx + φ2x = Fδ2(ρ) :=

1

(δ2 + f(ρ))2

φ(±1) = 0.

(17)

Our strategy is the following: we shall first replace the term f(ρ) by

f(ρ) :=

f(ρ) if ρ ∈ [0, 1]

0 otherwise

in order to have the right-hand side Fδ2(ρ) uniformly bounded and non-degenerate.The result is stated in the Lemmas 2.2 and 2.3. Then, we use the estimates on theelliptic coupling in order to prove that the density ρ satisfies ρ ∈ [0, 1], see Lemma

2.4. Since the solution to the f -modified system coincides with the one to (11), byuniqueness of smooth solutions to the regularized problem (11) we conclude that

the estimates for ρ and φ hold without replacing f by f . In order to simplify thenotation, we shall drop the tilde symbol above f .

Let us introduce the Hopf–Cole transformation

ψ(x, t) := e−φ(x,t)

δ1 , (18)

which implies

ψx = −ψφxδ1

, ψt = −ψφtδ1

, φx = −δ1ψxψ, φt = −δ1

ψtψ

(19)

ψxx = −φxxψδ1

− φxψxδ1

=ψ

δ21

(−δ1φxx + φ2

x

)=ψ

δ21

Fδ2(ρ). (20)

9

Therefore, ψ satisfies δ2

1ψxx = ψFδ2(ρ)

ψ(±1) = 1.(21)

As a first estimate, we prove that ψ is uniformly bounded in H1 and in L∞.

Lemma 2.2. There exists a constant C > 0 depending only on δ1 and δ2 such that

‖ψ‖H1([−1,1]) ≤ C, ‖ψ‖L∞([−1,1]) ≤ C, ‖ψxx‖L∞([−1,1]) ≤ C. (22)

Proof. Let us introduce the variable

ψ := ψ − 1,

which satisfies δ2

1ψxx = ψFδ2(ρ) + Fδ2(ρ)

ψ(±1) = 0.(23)

Multiplication of (23) by ψ and integration over [−1, 1] leads to (after integrationby parts)

−δ21

∫ψ2xdx =

∫ψ2Fδ2(ρ)dx+

∫ψFδ2(ρ)dx.

Since1

(1 + δ2)2≤ Fδ2(ρ) ≤ 1

δ22

, (24)

by a trivial use of Young’s inequality we get∫ψ2xdx+

∫ψ2dx ≤ C,

for a constant C depending on δ1 and δ2. Sobolev’s inequality then implies

‖ψ‖L∞ ≤ C.

The last assertion in (22) follows by the equation (21).

Next we prove that ψ is non-negative on [−1, 1] and uniformly bounded frombelow by a positive constant, which implies the desired estimates on the φ variable.

Lemma 2.3. There exists a constant C > 0 such that

ψ(x, t) ≥ C (25)

for all (x, t) ∈ [−1, 1]× [0,+∞). Moreover,

‖φ‖H1([−1,1]) ≤ C, ‖φ‖L∞([−1,1]) ≤ C, ‖φxx‖L∞([−1,1]) ≤ C. (26)

10

Proof. Let us consider the original equation (17) satisfied by φ. We have

δ1φxx +1

δ22

≥ δ1φxx + Fδ2(ρ) = φ2x ≥ 0,

which can be written as (δ1φ+

|x|2

2δ2

)xx

≥ 0.

Therefore the function δ1φ+ |x|2

2δ2attains its maximum at the boundary, φ is bounded

from above and ψ = e−φ/δ1 is bounded away from zero. The statements (26) followas a consequence of (25) and of (19)-(20).

We conclude by proving that ρ is always bounded above by the maximal densityρ = 1.

Lemma 2.4 (Boundedness of ρ). Assume that ρI ≤ 1. Then the solution to (16a)with f(ρ) = (1− ρ) satisfies ρ(x, t) ≤ 1 for all (x, t) ∈ [−1, 1]× [0,+∞).

Proof. We first define the function

η(ρ) =

0 ρ ≤ 0,ρ2

4γ0 < ρ ≤ 2γ,

ρ− γ ρ > 2γ.

(27)

and use it to approximate (ρ− 1)+ (the positive part of (ρ− 1)). Here γ > 0 is asmall parameter. Our goal is to show that this positive part, being zero at t = 0,does not increase. We consider

d

dt

∫η(ρ− 1) dx =

∫η′(ρ− 1)(ερx + (ρ(1− ρ)2φx)x dx

= −ε∫η′′(ρ− 1)ρ2

x dx+ εη′(ρ− 1)ρx|x=±1

−∫

0≤(ρ−1)≤γη′′(ρ− 1)ρ(1− ρ)2ρxφx dx+ η′(ρ− 1)ρ(1− ρ)φx|x=±1

≤ −2ε

∫η′′(ρ− 1)ρ2

x dx− Cε∫

0≤(ρ−1)≤γη′′(ρ− 1)ρ2(1− ρ)4|φx|2 dx

≤ −2ε

∫η′′(ρ− 1)ρ2

x dx− Cε,δγ3(1 + γ)2,

where Cε,δ depends on ε, δ1 and δ2. Here, we employed Young’s inequality andthe Dirichlet boundary conditions. Furthermore we used the ε-independent L∞

bound on φx we obtained in Lemma 2.3. Letting γ → 0, we infer

d

dt

∫(ρ− 1)+ dx = −ε

∫η′′(ρ− 1)|ρx|2 dx ≤ 0,

11

and thus the integral is decreasing in time. As (ρ− 1)+ is a positive function andzero at t = 0, we conclude that is stays zero for all times and thus that ρ is alwaysbounded by 1.

Note that using the same technique, but approximating the negative part of ρwe also obtain that the solution is almost everywhere non-negative (since ρI ≥ 0).

2.2. BV estimate on ρWe are now ready to prove the crucial BV estimate on ρ which serves as a

tool to get compactness in the limit as ε→ 0. Furthermore, it will guarantee theexistence of tr ρ, see [7, Lemma 1]. Let us start with estimating the L1 norm ofρx.

Lemma 2.5. Suppose ρI ∈ W 1,1([−1, 1]). Then, there exists a constant C > 0independent on ε such that

‖ρx(t)‖L1(Ω) ≤ (‖(ρI)x‖L1(Ω) + C)eCt

for all t ≥ 0.

Before we start the proof let us define an approximation ηγ(z) of the function|z| as γ → 0 such that

ηγ(z)→ |z|, η′γ(z)→ sign(z), η′γ(z)z → |z| as γ → 0

η′γ(z)z ≥ 0, η′′γ(z) ≥ 0

η′′γ(z) ≤ 1[−γ,γ](z)C

γ

(28)

for some constant C > 0.

Remark 2.6 (Properties of η). We remark that the definition of η implies thefollowing properties, which shall be often used in the sequel:

• All integrals of the form, with f ∈ L2(Ω),∫Ω

η′′γ(f(x))f(x)2 dx ≤ C

γ

∫|f(x)|≤γ

f(x)2 dx ≤ Cγ|Ω|

tend to zero as γ → 0.

• Furthermore, with g ∈ C1(R+), f, h ∈ L1(Ω), k ∈ R > 0 we have∫Ω

η′′γ(f(x)− k)(g(f(x))− g(k))h(x) dx

≤ C

γ

∫0<|f(x)−k|≤γ

‖g′‖L∞(R+)|f(x)− k||h(x)| dx

≤ C

γγ‖g′‖L∞(R+)

∫0<|f(x)−k|≤γ

|h(x)| dx→ 0,

12

as γ → 0.

Proof of Lemma 2.5. We deduce that

d

dt

∫ηγ(ρx)dx =

∫η′γ(ρx)ρxtdx =

∫η′γ(ρx)(g(ρ)φx)xxdx+ ε

∫η′γ(ρx)ρxxxdx

=

∫η′γ(ρx)(g

′(ρ)ρxφx)x +

∫η′γ(ρx)(g(ρ)φxx)x − ε

∫η′′γ(ρx)ρ

2xxdx

= −∫η′′γ(ρx)ρxxg

′(ρ)ρxφxdx+

∫η′γ(ρx)g

′(ρ)ρxφxxdx

+

∫η′γ(ρx)g(ρ)φxxxdx− ε

∫η′′γ(ρx)ρ

2xxdx

≤ −ε2

∫η′′γ(ρx)ρ

2xxdx+ C(ε)

∫η′′γ(ρx)φ

2xρ

2xdx+ C

∫|ρx|dx+ C. (29)

Here the last step is justified by the identities (19) and (20), by (25), and by

‖ψxxx(t)‖L1(Ω) ≤ C‖ρx(t)‖L1(Ω) + C, since ψxxx = Fδ2(ρ)ψx + ψF ′δ2(ρ)ρx.

The sum of the boundary terms∫η′γ(ρx)(ερxx + g′(ρ)ρxφx + g(ρ)φxx) dσx =

∫η′γ(ρx)ρt dσx

vanishes, as ρt is constant along the boundary. Due to Rem. 2.6, the second termon the right hand side of (29) vanishes as γ → 0, therefore we obtain the desiredassertion in the limit (after integration with respect to time).

Before estimating the L1 norm of ρt we have the following technical lemma.

Lemma 2.7. There exists a constant C > 0 independent of ε and of t such that

‖ψt(t)‖L∞(Ω) ≤ C‖ρt(t)‖L1(Ω) (30)

‖ψxxt(t)‖L1(Ω) ≤ C‖ρt(t)‖L1(Ω) (31)

‖ψxt(t)‖L∞(Ω) ≤ C‖ρt(t)‖L1(Ω). (32)

Proof. We start with the proof of estimate (30). Differentiation of (21) with respectto time yields

ψxxt =1

δ21

(ψtFδ2(ρ) + ψF ′δ2(ρ)ρt

). (33)

Next we multiply (33) by ψt and integrate over [−1, 1]. Using the fact that ψt = 0at the boundary, we integrate by parts to obtain

−δ21

∫ψ2xtdx =

∫Fδ2(ρ)ψ2

t dx+

∫F ′δ2(ρ)ρtψψtdx.

13

In view of (24) and Lemma 2.3 we can find a constant C = C(δ1, δ2) > 0 such that

‖ψt(t)‖2H1(Ω) ≤ C‖ψt(t)‖L∞(Ω)‖ρt(t)‖L1(Ω),

and the Sobolev inequality ‖ψt(t)‖L∞(Ω) ≤ ‖ψt(t)‖H1(Ω) implies the assertion.The inequality (31) follows by a direct use of the equation (21) and by (30).

Finally, the last statement (32) follows from the inequality

‖ψxt(t)‖L∞(Ω) ≤ ‖ψxxt(t)‖L1(Ω),

which is a consequence of the fact that∫ψxtdx = ψt(1, t) − ψt(−1, t) = 0 and

that every W 1,1 function in one space dimension admits an absolutely continuousrepresentative.

We are now ready to estimate the L1 norm of the time derivative.

Lemma 2.8. Assuming ρI ∈ W 2,1([−1, 1]) and ε > 0, there exists a constantC > 0 independent on ε such that

‖ρt(t)‖L1(Ω) ≤ CeCt,

for all t ≥ 0.

Proof. Again we consider the approximation ηγ of the absolute value, given by(28). We deduce that

d

dt

∫ηγ(ρt)dx =

∫η′γ(ρt)ρttdx =

∫η′γ(ρt)(g(ρ)φx)txdx+ ε

∫η′γ(ρt)ρxxtdx

=

∫η′γ(ρt)(g

′(ρ)ρtφx)x +

∫η′γ(ρt)(g(ρ)φxt)x − ε

∫η′′γ(ρt)ρ

2xtdx

= −∫η′′γ(ρt)ρxtg

′(ρ)ρtφxdx+

∫η′γ(ρt)g

′(ρ)ρxφxtdx

+

∫η′γ(ρt)g(ρ)φxxtdx− ε

∫η′′γ(ρt)ρ

2xtdx

≤ −ε2

∫η′′γ(ρt)ρ

2xtdx+ C(ε)

∫η′′γ(ρt)φ

2xρ

2tdx

+ C‖φxt(t)‖L∞(Ω)

∫|ρx|dx+ C

∫|φxxt|dx.

All boundary terms in the above calculation are zero as ρt and thus η′γ(ρt) is zeroon the boundary. The second term on the r.h.s. above vanishes as γ → 0. As forthe other terms, we can differentiate (18) to easily obtain

‖φxt(t)‖L∞(Ω) ≤ C‖ψxt(t)‖L∞(Ω) + C‖ψt(t)‖L∞(Ω) ≤ C‖ρt(t)‖L1(Ω)

14

and

‖φxxt(t)‖L1(Ω) ≤ C‖ψxxt(t)‖L1(Ω) +C‖ψxt(t)‖L1(Ω) +C‖ψt(t)‖L1(Ω) ≤ C‖ρt(t)‖L1(Ω).

Therefore, integration with respect to time and Lemma 2.5 results in

‖ρt(t)‖L1(Ω) ≤ (‖(ρ(0)t‖L1(Ω) + C)eCt,

for all t ≥ 0. Using the fact that ρI is in W 2,1(Ω) and that ε is bounded, we canuse equation (16a) to estimate

‖ρt(0)‖L1(Ω) ≤ ‖g′(ρI)φx(t)‖L∞(Ω)‖(ρI)x‖L1(Ω) + ε‖(ρI)xx‖L1(Ω).

We thus conclude that ‖ρt(0)‖L1(Ω) is bounded as well completing the proof.

2.3. Stability estimates on φ

Next, we prove some stability estimates for the elliptic equation (17) with re-spect to the variable ρ. These estimates will be useful later on to prove uniquenessof an entropy solution ρ in the limit.

Given two densities ρ and ρ, let φ and φ solve

− δ1φxx + φ2x = Fδ2(ρ),

− δ1φxx + φ2x = Fδ2(ρ),

with boundary conditions φ(±1) = φ(±1) = 0. For both solutions we consider thecorresponding Hopf–Cole transformation

ψ(x, t) := e−φ(x,t)

δ1 ψ(x, t) := e− φ(x,t)

δ1 .

Then we can deduce the following lemma:

Lemma 2.9. There exists a constant C > 0 independent on ε and on t such that

‖φ(t)− φ(t)‖L1(Ω) ≤ C‖ρ(t)− ρ(t)‖L1(Ω) (34)

‖φxx(t)− φxx(t)‖L1(Ω) ≤ C‖ρ(t)− ρ(t)‖L1(Ω) (35)

‖φx(t)− φx(t)‖L∞(Ω) ≤ C‖ρ(t)− ρ(t)‖L1(Ω). (36)

Proof. Let us multiply equation

δ21(ψxx − ψxx) = (ψ − ψ)Fδ2(ρ) + ψ(Fδ2(ρ)− Fδ2(ρ)) (37)

by η′γ(ψ− ψ), with ηγ given by (28) and integrate over [−1, 1]. Integration by partsimplies

−δ21

∫(ψx − ψx)2η′′γ(ψ − ψ)dx

=

∫(ψ − ψ)η′γ(ψ − ψ)Fδ2(ρ)dx+

∫ψη′γ(ψ − ψ)[Fδ2(ρ)− Fδ2(ρ)]dx.

15

We use the properties of ηγ and (24) to obtain, as γ → 0

C(δ)

∫|ψ − ψ|dx ≤

∫Fδ(ρ)|ψ − ψ|dx ≤

∫ψ|Fδ(ρ)− Fδ(ρ)|dx ≤ C

∫|ρ− ρ|dx.

Next we can deduce (34) by using the Hopf–Cole transformation as usual. Toprove (35), multiply (37) by sign(ψxx − ψxx) and integrate over [−1, 1] to obtain

δ21

∫|ψxx − ψxx|dx ≤ C

∫|ψ − ψ|dx+ C

∫|ρ− ρ|dx.

Next we obtain (35) by using (34) and passing to the variable φ. Inequality (36)follows by the Sobolev inequality as at the end of the proof of Lemma 2.7.

2.4. The limit as ε→ 0

Our next goal is to study the behaviour of the solution (ρε, φε) to the system(16) as the parameter ε tends to zero. Using Lemma 2.5 and Lemma 2.8 we knowthat ρε is in the space of functions having bounded variation BV (Ω). Therefore,we can employ the classical Helly’s theorem on strong L1–compactness of functionswith bounded BV–norm, cf. [11] for instance. Thus, ρε has a strong limit in L1

up to subsequences. As for the φ variable, since ρx is uniformly estimated in L1,differentiating the elliptic equation with respect to x implies that φεxxx is uniformlybounded in L1 and therefore φεxx is strongly compact in L1. Denoting by (ρ, φ) thelimit ε→ 0 of (ρε, φε), as the convergence is strong in L1 and due to the estimateson φ proven in subsection 2.1, it is immediately clear that φ solves the secondequation in (11) and ρ is a weak solution of

ρt − (ρf 2(ρ)φx)x = 0. (38)

In the remainder of this section, we will show that (ρ, φ) is in fact the uniqueentropy solution to the system (11) in the sense of Definition 2.1. First we shallstate the existence theorem.

Theorem 2.10 (Existence of entropy solutions). There exists an entropy solution(ρ, φ) to system (11) with initial condition (12) and boundary conditions (13)-(14)in the sense of Definition 2.1. Such solution is the limit as ε → 0 of the solutionρε to (16a)-(16b).

Proof. To recover the notion of entropy solutions, we consider again the regularizedequation

ρt = (ρf 2(ρ)φx)x + ερxx. (39)

16

We multiply this equation by η′(ρ − k)ψ (with η′ defined in (28)) and integrateover ΩT = [−1, 1]× [0, T ]∫∫

ΩT

η′(ρ− k)ρtψ dxdt =

∫∫ΩT

η′(ρ− k)(g(ρ)φx)xψ dxdt

+ ε

∫∫ΩT

η′(ρ− k)ρxxψ dxdt.

Adding

0 =

∫∫ΩT

η′(ρ− k)g(k)φxψx dxdt−∫∫ΩT

η′(ρ− k)g(k)φxψx dxdt

and integrating by parts leads to∫∫ΩT

η′(ρ− k)ρtψ dxdt = −∫∫ΩT

η′(ρ− k)[g(ρ)− g(k)]ψxφx dxdt

+

∫∫ΩT

η′(ρ− k)g(k)φxxψ dxdt−∫∫ΩT

η′′(ρ− k)[g(ρ)− g(k)]φxρxψ dxdt

−∫ T

0

η′(k)(g(0)− g(k))φxψ|x=±1dt− ε∫∫ΩT

η′′(ρ− k)ρ2xψ dxdt

−ε∫∫ΩT

η′(ρ− k)ρxψx dxdt+

∫ T

0

εη′(ρ− k)ρxψ|x=±1dt

≤ −∫∫ΩT

η′(ρ− k)[g(ρ)− g(k)]ψxφx dxdt

+

∫∫ΩT

η′(ρ− k)g(k)ψφxx dxdt− η′(k)

∫ T

0

[g(0)− g(k)]φxψ|x=±1dt

−ε∫∫ΩT

η(ρ− k)ρxψx dxdt− η′(k)

∫ T

0

ερxψ|x=±1dt

−∫∫ΩT

η′′(ρ− k)[g(ρ)− g(k)]φxρxψ dxdt

Next we integrate the first term by parts and multiply it by −1. Taking the limitas γ → 0 the last term on the right hand side vanishes (due to the continuity of g

17

and the boundedness of φx and ψ, cf. Remark 2.6) and we obtain∫∫ΩT

|ρ− k|ψt dxdt+

∫ 1

−1

ρI(x)ψ(x, 0) dx

≥∫∫ΩT

sgn(ρ− k)[g(ρ)− g(k)]ψxφx dxdt

−∫∫ΩT

sgn(ρ− k)g(k)ψφxx dxdt+ sgn(k)

∫ T

0

[g(0)− g(k)]φxψ|x=±1dt

+ ε

∫∫ΩT

|ρ− k|ρxψx dxdt+ sgn(k)

∫ T

0

ερxψ|x=±1dt.

(40)

Next we consider the limit ε→ 0. Using Lemma 2.5, the fourth term on the righthand side can be estimated by∣∣∣∣∣∣ε

∫∫ΩT

|ρ− k|ρxψx dxdt

∣∣∣∣∣∣ ≤ εC‖ψx(t)‖L∞(Ω), (41)

and thus tends to zero. To compute the limit for the last term, i.e.

limε→0

ε

∫ T

0

ρxψ|x=±1dt,

we introduce (following [7]), for some κ > 0 the function ξκ ∈ C2([−1, 1]) with thefollowing properties

ξκ(x) = 1 on x = ±1,ξκ(x) = 0 on x ∈ [−1, 1] ; dist(x, ∂[−1, 1]) ≥ κ ,

0 ≤ ξκ(x) ≤ 1 on (−1, 1).(42)

Furthermore, defining M([−1, 1]) as the space of Radon measures on [−1, 1], wechoose ξκ such that

∂xξκ → µ|−1,1 ∈M([−1, 1]) as κ→ 0,

defined as

µ = δx=1 − δx=−1.

Now we obtain

ε

∫∫ΩT

ρxxψξκ dxdt = −ε∫∫ΩT

ρx(ψξκ)x dxdt+ ε

∫ T

0

ρxψ|x=±1dt.

18

The second term in this equation

−ε∫∫ΩT

ρx(ψξκ)x dxdt = −ε∫∫ΩT

ρx(ψxξκ + φ(ξκ)x) dxdt

vanishes in the limit ε → 0 due to the L∞ bounds on ψ, ψx, ξκ, (ξκ)x (givenfor κ > 0 since ξκ ∈ C2([−1, 1])) and the L1-boundedness of ρx. Using (39) wetherefore obtain

limε→0

(ε

∫ T

0

ρxψ|x=±1dt

)= −

∫∫ΩT

(ρψt − g(ρ)φxψx) ξκ dxdt+

∫Ω

ρψξk dx

∣∣∣∣t=0

+

∫∫ΩT

g(ρ)φxψ(ξκ)x dxdt−∫ T

0

g(0)φxψ|x=±1dt.

Finally letting κ→ 0, the first term on the right hand side tends to zero while thesecond tends to an evaluation on the boundary. Due to the continuity of ρ and ψthe boundary term resulting from the integration by parts in time vanishes as thesupport of ξκ converges to a set of Lebesgue measure zero (i.e. −1, 1). Thus wehave

limε→0

ε

∫ T

0

ρxψ|x=±1dt =

∫ T

0

(g(tr ρ)− g(0))φx(s, t)ψ|x=±1dt.

Combining this result with (40) we finally obtain the entropy formulation as inDefinition 2.1 and this completes the proof.

Next we prove that the boundary condition (13) can be recovered by the defi-nition of entropy solution.

Lemma 2.11. Let ρ be an entropy solution given by Definition 2.1. Then, thefollowing inequality holds for all k ∈ [0, tr ρ]

g(tr ρ) ≥ g(k) at x = ±1. (43)

Proof. In (15), we choose the special test function ψ = ν(t)ωκ with ν ∈ C2(]0, T [)positive and ωκ ∈ C2([−1, 1]) with the following properties:

ωκ(x) = 1 on x = −1,ωκ(x) = 0 on x ∈ [−1, 1] ; |x+ 1| ≥ κ ,

0 ≤ ωκ(x) ≤ 1 on (−1, 1).(44)

Similarly as before for ξκ, we choose ωκ such that

∂xωκ → −δ−1 as κ→ 0,

19

where δ−1 denotes the Dirac delta measure centered at −1. Then, in the limitκ→ 0 (15) converges to∫ T

0

sgn(tr ρ− k)[g(tr ρ)− g(k)]φx|x=−1ν(t)dt

+ sgn(k)

∫ T

0

[g(tr ρ)− g(k)]φx|x=−1ν(t)dt ≥ 0,

for all k ∈ R. Thus, almost everywhere in −1 × (0, T ) we have

(sgn(tr ρ− k) + sgn(k))[g(tr ρ)− g(k)]φx ≥ 0.

To conclude the proof we note that φx is always (i.e. independently of the givenρ) non-negative at x = −1. This is a consequence of the fact that φ = 0 atx = ±1 (boundary conditions) and positive on the whole domain, due to a trivialminimum principle for the equation (16b). Employing Hopf’s Lemma we thereforeconclude strict positivity of φx at x = −1. In a similar way, one can construct afunction ωk concentrating on x = 1 with a derivative converging to a Dirac delta atx = 1. The same inequality is obtained since the change of sign in the derivative ofconcentrator ωk is balanced by the change of sign in φx (non-increasing at x = 1).To conclude, we note that (sgn(tr ρ − k) + sgn(k)) = 0 for all k /∈ [0, tr ρ] (astr ρ ≥ 0) and equal to 2 otherwise.

2.5. Uniqueness

Next we shall prove that the entropy solution in the sense of Definition 2.1 isunique.

Theorem 2.12 (Uniqueness of entropy solutions). There exists at most one en-tropy solution (ρ, φ) to the system (11) with initial condition (12) and boundaryconditions (13)-(14) in the sense of Definition 2.1.

The above stated result is a consequence of the following stability theorem,which follows the same technique developed in [19]. Here the authors use thevariables doubling technique originally introduced in [22]. A similar strategy wasalso used e.g. [3, 4].

We state the following useful result:

Lemma 2.13. ([19]) Consider a function z = z(x) belonging to L∞(Rd)∩BV (Rd)and let h be Lipschitz on the interval Iz := [−‖z‖L∞ , ‖z‖L∞ ]. Then h(z) belongsto L∞(Rd) ∩BV (Rd) and ∣∣∣∣ ∂∂xj h(z)

∣∣∣∣ ≤ ‖h‖Lip(Iz)

∣∣∣∣ ∂∂xj z∣∣∣∣

in the sense of measures for j = 1, . . . , d.

20

Uniqueness can be deduced from the following theorem:

Theorem 2.14. Let (ρ, φ), (ρ, φ) be the two entropy solutions to system (11)according to Definition 2.1 with initial data ρI , ρI ∈ L∞([−1, 1]) ∩ BV ([−1, 1])respectively. Then for almost all t ∈ (0, T ),

‖ρ(t)− ρ(t)‖L1(Ω) ≤ ‖ρI − ρI‖L1(Ω) + t‖g‖L∞(Ω)‖φxx(t)− φxx(t)‖L∞((0,T );L1(Ω))

+ t‖g‖Lip(Ω)‖ρx(t)‖L1(Ω)‖φx(t)− φx(t)‖L∞((0,T );L∞(Ω))

holds.

Combining this result with (35) and (36) from Lemma 2.9 we obtain

‖ρ(t)− ρ(t)‖L1(Ω) ≤ ‖ρI − ρI‖L1(Ω) + tC‖ρ(t)− ρ(t)‖L1(Ω), (45)

for some positive constant C. Choosing t small enough this inequality contradictsthe existence of two different solutions ρ and ρ having the same initial datum andthus implies uniqueness. It remains to prove Theorem 2.14.

Proof. We first note that in this proof there will sometimes, after integration byparts, be terms which insolve derivatives of the sgn. To be precise, the sgn needsto be approximated in these situations, as in the proof of Lemma 2.5. However,to increase the readability of this proof, we will omit this detail here. Considera nonnegative, compactly supported, Lipschitz continuous function ψ(x, t, x, t),defined on [−1, 1]× [0, T [×[−1, 1]× [0, T [. Furthermore, let ψ be zero on −1, 1×[0, T ). Next, we take two admissible solutions ρ(x, t), ρ(x, t) and write (15) as∫∫ΩT

|ρ− ρ|ψt dxdt−∫∫ΩT

sgn(ρ− ρ)[g(ρ)− g(ρ)]ψxφx(x, t) dxdt +

∫∫ΩT

sgn(ρ− ρ)g(ρ)ψφxx(x, t)dxdt− sgn(ρ)

∫ T

0

[g(tr ρ)− g(ρ)]φx(x, t)ψ |x=±1 dt ≥ 0.

and∫∫ΩT

|ρ− ρ|ψt dxdt−∫∫ΩT

sgn(ρ− ρ)[g(ρ)− g(ρ)]ψxφx(x, t) dxdt +

∫∫ΩT

sgn(ρ− ρ)g(ρ)ψφxx(x, t) dxdt− sgn(ρ)

∫ T

0

[g(tr ρ)− g(ρ)]φx(x, t)ψ |x=±1 dt ≥ 0.

Integrating both the above inequalities over ΩT := Ω× [0, T [, the first with respectto x, t and the second with respect to x, t and adding the resulting equations leads

21

to∫∫∫∫ΩT×ΩT

|ρ− ρ|(ψt + ψt) dzdz

−∫∫∫∫ΩT×ΩT

[sgn(ρ− ρ)

(g(ρ)φx(x, t)− g(ρ)φx(x, t)

)(ψx + ψx)

]︸ ︷︷ ︸:=I1

dzdz

−∫∫∫∫ΩT×ΩT

[sgn(ρ− ρ)

(g(ρ)ψx

(φx(x, t)− φx(x, t)

)+ g(ρ)ψx

(φx(x, t)− φx(x, t)

))]︸ ︷︷ ︸:=I2,1

dzdz

+

∫∫∫∫ΩT×ΩT

[sgn(ρ− ρ)(g(ρ)φxx(x, t)− g(ρ)φxx(x, t))ψ

]︸ ︷︷ ︸:=I2,2

dzdz

=

∫∫∫∫ΩT×ΩT

(|ρ− ρ|(ψt + ψt) + I1 + I2,1 + I2,2) dzdz ≥ 0.

Here z = (x, t) and z = (x, t). We take a symmetric function δ ∈ C∞(R) with totalmass one and Supp(δ) ⊂ (−1, 1). We define

δh(·) :=1

hδ( ·h

)and choose the following test function

ψ = ν

(t+ t

2,x+ y

2

)δh

(t− t

2

)δh

(x− x

2

).

From this definition we conclude∫∫∫∫ΩT×ΩT

(|ρ− ρ|(ψt + ψt) + I1) dxdtdxdt

=

∫∫∫∫ΩT×ΩT

(|ρ− ρ|νt + sgn(ρ− ρ)

(g(ρ)φx(x, t)− g(ρ)φx

)νx)×

× δh(t− t

2

)δh

(x− x

2

)dxdtdxdt.

We now consider the term I2,1

I2,1 = − sgn(ρ− ρ)[φx(x, t) (g(ρ) + g(ρ))− φx(x, t) (g(ρ) + g(ρ))

] 1

2νxδhδh

− sgn(ρ− ρ)[φx(x, t) (g(ρ)− g(ρ))− φx(x, t) (g(ρ)− g(ρ))

]ν(δhδh)x

=: I2,1,1 + I2,1,2.

22

Here, we used that by definition we have νx = 12νx and (δhδh)x = −(δhδh)x. Inte-

grating by parts in I2,1,2 leads to

−∫∫∫∫ΩT×ΩT

sgn(ρ− ρ)[φx(x, t) (g(ρ)− g(ρ))− φx(x, t) (g(ρ)− g(ρ))

]×

× ν(δhδh)x dxdtdxdt

=

∫∫∫∫ΩT×ΩT

sgn(ρ− ρ)[φx(x, t) (g(ρ)− g(ρ))− φx(x, t) (g(ρ)− g(ρ))

]×

× 1

2νxδhδh dxdtdxdt

+

∫∫∫∫ΩT×ΩT

φx[(sgn(ρ− ρ)(g(ρ)− g(ρ)))x − φxx(x, t) sgn(ρ− ρ) (g(ρ)− g(ρ))]×

× νδhδh dxdtdxdt

+

∫∫∫∫ΩT×ΩT

− φx(x, t)(sgn(ρ− ρ) (g(ρ)− g(ρ)))xνδhδh dxdtdxdt.

Noticing that

−φxx(x, t) sgn(ρ− ρ) (g(ρ)− g(ρ)) + I2,2

= − sgn(ρ− ρ)(φxx(x, t))− φxx(x, t))g(ρ)νδhδh

and adding again I2,1,1 we obtain∫∫∫∫ΩT×ΩT

(I2,2 + I2,1,2 + I2,1,1) dxdtdxdt

=

∫∫∫∫ΩT×ΩT

− sgn(ρ− ρ)(φxx(x, t)− φxx(x, t))g(ρ)νδhδh dxdtdxdt

+

∫∫∫∫ΩT×ΩT

(φx − φx)(sgn(ρ− ρ)(g(ρ) + g(ρ)))xνδhδh dxdtdxdt

+

∫∫∫∫ΩT×ΩT

sgn(ρ− ρ)[φx(x, t)g(ρ)− φx(x, t)g(ρ)

]νxδhδh)︸ ︷︷ ︸

=:J

dxdtdxdt.

As there are no more derivatives in the terms involving δhδh, we consider the limith→ 0, remove two integrals and set x = x, t = t. This is a rather technical pointwhich is explained in great detail in [21]. We choose the new test function

ν(x, t) = νκ,h(x, t) = (1− ξκ(x))χh(t),

23

with for some 0 < t1 < t2 < T fixed

χh(t) =

∫ t

−∞(δh(τ − t1)− δh(τ − t2)) dτ,

and ξκ as defined in (42). We observe that all terms which are bounded in L1 andmultiplied by (νκ,h(x, t))x converge to a boundary term in the limit κ → 0. Wethus have

limh→0κ→0

∫∫ΩT

(I1 + J) dxdt = −∫ t2

t1

∫∂Ω

sgn(tr ρ− tr ρ)φx[g(tr ρ)− g(tr ρ)] dsdt,

and therefore

− limh→0,κ→0

∫∫ΩT

(|ρ− ρ|νt + I1 + I2,1 + I2,2) dxdt

= −∫ 1

−1

(|ρ− ρ|) dx∣∣∣∣t2t1

+

∫ t2

t1

∫ 1

−1

− sgn(ρ− ρ)(φxx(x, t))− φxx(x, t))g(ρ) dxdt

+

∫ t2

t1

∫ 1

−1

(φx − φx)(sgn(ρ− ρ)(g(ρ) + g(ρ)))x dxdt

+

∫ t2

t1

sgn(tr ρ− tr ρ)φx[g(tr ρ)− g(tr ρ)]|x=±1dt ≥ 0.

Using Lemma 2.13, we have

|(sgn(tr ρ− tr ρ)(g(tr ρ)− g(tr ρ)))x| ≤ ‖g‖Lip(I)|ρx|. (46)

Collecting all the above terms we obtain

‖ρ(t)− ρ(t)‖L1(Ω)

∣∣t2t1≤∫ t2

t1

∫Ω

[|φxx(x, t)− φxx(x, t)|‖g‖L∞(Ω)

+‖φx(t)− φx(t)‖L∞(Ω)‖g‖Lip(I)|ρx|]dxdt

+

∫ t2

t1

∫∂Ω

sgn(tr ρ− tr ρ)φx[g(tr ρ)− g(tr ρ)] dsdt.

(47)

Following [7], we define

k(x, t) =

tr ρ if tr ρ < tr ρ,0 if tr ρ = tr ρ,

tr ρ if tr ρ > tr ρ.

24

This allows us to write, at x = −1

sgn(tr ρ− tr ρ)φx(−1, t)[g(tr ρ)− g(tr ρ)] = sgn(tr ρ− k)φx(−1, t)[g(tr ρ)− g(k)]

+ sgn(tr ρ− k)φx(−1, t)[g(tr ρ)− g(k)].

Note that φx(−1, t) > 0. At x = 1, the same holds true. Using Lemma 2.11 weconclude that the last term on the right hand side of (47) is negative and cantherefore be omitted. Thus letting t1 → 0 we arrive at the desired inequality andthis completes the proof.

Remark 2.15 (Alternative Regularization). We also considered the alternativeregularisation as introduced in Sec. 1.2, namely system (16) with δ2 = 0. We wereable to show the existence of weak solutions. However, it was not possible to obtaina uniqueness result. Even though we still obtained a bound on ‖ρx(t)‖L1(Ω) (using aAubin-Lions like argument) we couldn’t obtain this bound for ρt. The main reasonfor this is that the Hopf-Cole transform cannot be used for this regularisation andit was not possible to control terms of the form φxt and φxxt. Detailed results willbe part of the PhD thesis [33].

3. Numerics and Examples for the Hughes’ model

In this section we discuss the behaviour of solutions for the non regularizedone-dimensional problem with simple initial data. Already these examples showquite interesting features which can be reproduced by numerical simulations. Thecontent of this section is formal as we don’t provide any existence and uniquenesstheory. However, the characteristic calculus provides a useful tool to understandqualitatively the behaviour of the solution in the simple examples considered andis in complete agreement with the numerical results.

3.1. Characteristic Calculus

We consider the non-regularized problem

ρt − (ρf 2(ρ)φx)x = 0, (48a)

|φx| =1

f(ρ). (48b)

In the following, we always consider the unique viscosity solution φ to (48b).We use (in a non rigorous way) the notion of a viscosity solution to be able tointerpret φ as a biased shortest distance to the exit. Note that thus this solutionhas a unique turning point x0(t) (i.e. point, where φx changes sign) given by theimplicit relation ∫ x0(t)

−1

1

f(ρ)dx =

∫ 1

x0(t)

1

f(ρ)dx.

25

Thus, (48a) can be written as (using that |φx| = φx sgnφx)

ρt − (ρf(ρ) sgnφx)x = 0. (49)

The natural boundary conditions (in the spirit of [7, 9]) are given by

f(tr ρ) ≥ f(k) on x = ±1, for all k ∈ [0, tr ρ], (50)

which is satisfied if and only if tr ρ belongs to the interval of densities correspondingto outgoing characteristics, i. e. tr ρ ∈ [0, 1/2]. As shown in [9], the boundarycondition in case of incoming characteristics is determined by solving a Riemannproblem between the boundary datum (i.e. zero in this case) and the trace of thedensity next to the boundary.

Away from the time dependent interface x = x0(t) (where φx is discontinuous)we can give sense to characteristics. They are defined by

x = −(1− 2ρ) sgn(φx).

Note that the Rankine-Hugoniot condition for a hyperbolic conservation law withflux F , i.e. ρt + F (ρ)x = 0 is given by

[[F (ρ)]] = x0(t) [[ρ]] . (51)

Here, [[·]] denotes the jump at the discontinuity x0.

3.1.1. Constant initial data

We would like to understand the behaviour of the solution in the very simplecase of constant initial data. Here we are particularly interested in the three caseswhich correspond to different characteristic speeds, i.e. ρI less, equal or greaterthan 1/2. In particular we consider the cases ρI = 1/4, ρI = 1/2 and ρI = 3/4. Inthe case of constant initial data, the interface is constant in time, i.e. x0 = 0 andlocated at x = 0. Thus sgnφx = − sgnx and (48a) can be written as

ρt + (ρf(ρ) sgnx)x = 0. (52)

The RH condition (51) for this flux F (ρ) = ρf(ρ) sgnx reads

f(ρ+) + f(ρ−) = 0,

where ρ± denote the right and left limit of ρ at the interface x = 0. An immediateconsequence of this is that constant functions ρ(x, t) = c with c ∈ (0, 1) do notsatisfy the RH condition (51) and are not weak solutions. If we start with aconstant initial datum we expect the equation to “correct” this by forcing ρ(0, t) =0 in arbitrary small time (ρ(x, t) = c would also create a solution, which however

26

a) b) c)

Figure 2: a) ρ = 0.25 b) ρ = 0.5 c) ρ = 0.75

does not fulfil the entropy condition). Then two shocks originate between ρ(0, t) =0 and ρ(x, t) = c for x 6= 0, which move towards the boundary. The slope of theseshocks is determined by the RH condition (51). In the three cases considered weobtain

x =

±3

4ρI(x) = 1

4

±12

ρI(x) = 12

±14

ρI(x) = 34.

This situation, locally around x = 0, is sketched in Fig. 2. Around the centerx = 0 where no information is transported to, we expect the solution to be eitherzero or a rarefaction wave. In case of a rarefaction wave we make the ansatzρ(x, t) = u

(xt

)and deduce from (52) that

uRF(x, t) =x+ t

2t.

This solution continuously connects the two outgoing shocks but creates the con-stant value 1/2 at x = 0 and is thus not admissible. Therefore, the we expectformation of a vacuum in between the two shocks in all three cases. In the caseρ = 3/4, we encounter an additional phenomenon at the boundaries. Here thecharacteristics point inwards, therefore we need to prescribe boundary conditionsat x = ±1. We choose the following Dirichlet boundary conditions ρ(±1, t) = 1/2(maximal flux). Such condition is easily recovered by solving the Riemann problembetween tr ρ = 3/4 and the boundary value zero (cf. [9]).

This implies that the characteristics at the boundary are vertical while char-acteristics of slope 1/2 transport the value 3/4 into the domain. Hence we obtaintwo wedges (one at each boundary) in which no information is transported bycharacteristics. If we make again the ansatz ρ(x, t) = u

(x+1t

)(shifted to the left

boundary), we obtain the following rarefaction wave

ρ(x, t) =x+ 1 + t

2t,

27

which is an admissible solution. Thus we expect rarefaction waves at both bound-aries. At time t = 4/3, these rarefaction waves will hit the shocks coming fromthe interface (at x = ±1/3, respectively). To calculate the new slope of the shockwe use the RH condition (51) which results in the following ODE

s(t) = −s(t)2t

+t− 1

2t, s

(4

3

)= −1

3.

Using standard techniques we obtain the solution

s(t) = −√t

(1 + t√t−√

3

).

A complete picture of the case ρI(x) = 3/4 is given in Fig. 3. In the next sectionwe will see that all these phenomena can be observed in numerical simulations.

t

x

t=43

Figure 3: Details for the case ρI(x) = 3/4

Remark 3.1 (Boundary conditions in the regularized and non regularized case).At a first glance there is a clear discrepancy between the boundary conditions inthe regularized case (13) and the ones prescribed above for the non regularizedmodel. In the latter case, the set of admissible boundary data is determined viathe monotonicity of f at the boundary, whereas in the former case this set isdetermined via the function g. Hence, there is the possibility of a boundary layerin a possible limit as δ1 → 0. However, the regularized problem has a source termg(ρ)φxx, and this fact could possibly imply some compensation phenomena at theboundary which can avoid the boundary layer. This issue will be the topic offuture study.

28

3.2. Numerical simulations

Next we present numerical simulations of (48) relating the results to the pre-vious discussion in Section 3.1. We consider the regularized system on the domainΩ = [−1, 1]

ρt − div(ρf(ρ) sgnφx) = ερxx (53a)

|φx| =1

f(ρ)(53b)

with a regularization parameter ε ≥ 0. The system is supplemented with the ini-tial condition ρ(x, 0) = ρI(x) and inhomogeneous Dirichlet boundary conditionsρ(±1, t) = ρD. We use these boundary conditions to be consistent with the char-acteristic calculus presented in Sec. 3.1. This allows us to compare the numericalresults with these computations. We solve (53) in an iterative manner, i.e.

1. Given ρ solve the eikonal equation (53b) with fast sweeping method.

2. Solve the non-linear conservation law (53a) for a given φ using an ENOscheme or resp. a Godunov scheme.

We choose the following discretisation. The domain R is divided into cellsIj = [xj− 1

2, xj+ 1

2] with centers at points xj = j∆x for j ∈ Z. The time domain

(0,∞) is discretised in the same manner via tn = n∆t resulting in time stripsIn = [tn, tn+1].

We used two different schemes to compare and understand the behaviour ofsolutions. In the first approach we use an ENO scheme with small diffusion onthe whole domain Ω = [−1, 1]. In the second approach we split the domain intotwo parts Ω = Ω1 ∪ Ω2 where Ω1 = [0, x(t)] and Ω2 = [x(t), 1], solve equation(53a) with a Godunov scheme (and no diffusion, i.e. ε = 0) on Ω1 and Ω2 andconcatenate both solutions.

3.2.1. ENO scheme

J. Towers presented convergence results for an ENO scheme for conservationlaws with discontinuous flux in [37]. This ansatz can be used in Step (2) to solve(53a) with small diffusion on the whole domain Ω = [−1, 1]. Let χnj denote thecharacteristic function on the rectangle Rn

j = Ij × In. The finite difference schemethen generates for every mesh size ∆x and ∆t a piecewise constant solution ρ∆

given by

ρ∆(x, t) =∑n≥0

∞∑−∞

χnj ρnj .

29

The approximations ρnj are generated by an explicit algorithm

ρn+1j = ρnj − λ1(kj+ 1

2hj+ 1

2− kj− 1

2hj− 1

2) + λ2(dj+ 1

2− dj− 1

2). (54)

Here λ1 = ∆t∆x

, λ2 = ε∆t∆x2 and kj± 1

2= sgnφx(xj± 1

2). The diffusive flux is given

by dnj+ 1

2

:= ρnj+1 − ρnj , the convective one hj+ 12

:= h(v, u) is chosen such that it

is consistent with the actual flux, i.e. h(ρ, ρ) = g(ρ) = ρf(ρ). To guaranteemonotonicity the flux is transposed when kj+ 1

2changes sign, i.e.

hj+ 12

=

h(ρj+1, ρj) if kj+ 1

2≥ 0

h(ρj, ρj+1) if kj+ 12< 0.

We choose the ENO flux [10] which is given by

h(v, u) =1

2(g(u) + g(v)) +

1

2

∫ v

u

|gu|du. (55)

Godunov scheme. The Godunov scheme is derived by using the exact solutionoperator for ρt + (F (ρ))x = 0 with piecewise constant initial data. The resultingnumerical flux is h(v, u) = F (uG(v, u)), where uG(v, u) is the similarity solutionof the resulting Riemann problem with right and left state v and u evaluatedanywhere on the vertical half-line t > 0 where the jump in the initial data occurs.The Godunov flux [31] is given by

h(v, u) =

min[u,v] F (w) if u ≤ v

max[u,v] F (w) if u ≥ v.(56)

Constant initial data. First we would like to validate the characteristic calculuspresented in section 3.1. We choose constant initial data ρI(x) that is smaller orlarger than 1/2. The time discretisation is set to ∆t = 10−4, the spatial discreti-sation to ∆x = 10−2. Here we solved the non regularized problem with ε = 0using Godunovs’ method. First we choose ρI(x) = 1/4, the evolution is depictedin Figure 4. In this case the characteristics point outward, therefore we prescribenumerical boundary conditions instead of physical ones. In our second example weset ρI(x) = 3/4. Here we observe a good agreement of the numerical simulationwith the theoretical results, see Figure 5. Note that the shock hits the rarefactionwaves at t = 4/3 and that the subsequent shock hits the boundary at t = 3 (aspredicted by our characteristic calculus).

30

(a) t = 14 (b) t = 3

4 (c) t = 54

Figure 4: Evolution of ρ with initial datum ρI(x) = 0.25

(a) t = 12 (b) t = 4

3 (c) t = 52

Figure 5: Evolution of ρ with initial datum ρI(x) = 0.75 and Dirichlet boundary conditionsρ(±1) = 0.5

Other examples. Finally we would like to illustrate the behavior with other exam-ples. We choose the following initial guess

ρI(x) =

0.8 if − 0.8 ≤ x ≤ −0.5

0.6 if − 0.3 ≤ x ≤ 0.3

0.9 if 0.4 ≤ x ≤ 0.75,

representing three groups which would like to exit at x = 1 or x = −1. We setthe spatial discretisation to ∆x = 10−3, the discretisation in time to ∆t = 10−4.Here we solve (53a) on the whole domain using an ENO flux and ε = 10−4. Theevolution of the densities is illustrated in Figure 6. Here the y axis correspondsto time, running from 0 (top) to 1.5 (bottom). The right group (located between0.4 ≤ x ≤ 0.75) splits at the beginning, a small part moves to the left while therest moves towards the right exit. We observe that the part of the group whichwas moving to the left changes direction and moves towards the right.

31

−1−0.8−0.6−0.4−0.200.20.40.60.81

0

0.5

1

1.5

Tim

e t

Space x

Crowd motion in time

(a) ρ(x, t)

0 0.5 1 1.5−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time t

x(t)

Evolution of x(t)

(b) x(t)

Figure 6: Evolution of ρ and x(t)

Acknowledgements

This publication is based on work supported by Award No. KUK-I1-007-43,made by King Abdullah University of Science and Technology (KAUST), by theLeverhulme Trust through the research grant entitled Kinetic and mean field par-tial differential models for socio-economic processes (PI Peter Markowich) and bythe Royal Society through the Wolfson Research Merit Award of Peter Markowich.PM is also grateful to the Humboldt foundation for their support. MDF is partiallysupported by the Italian MIUR under the PRIN program ’Nonlinear Systems ofConservation Laws and Fluid Dynamics’. Furthermore, the authors thank Mar-tin Burger and the Institute for Computational and Applied Mathematics at theUniversity of Munster for their kind hospitality and stimulating discussions.

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